Conditional Probability of defective bulb

I say urgent because of the horribly small lecture I received on this section, a whole 3 minutes or so of examples. While I won't give further context I can say without a doubt I am completely lost.

Here is the problem I am stuck on.

In a string of 12 Christmas tree light bulbs, 3 are defective. The bulbs are selected at random and tested, one at a time, until the third defective bulb is found. Compute the probability that the third defective bulb is the:

a) Third Bulb Tested
b) Fifth Bulb Tested
c) Tenth Bulb Tested

I am at a lost. My only feeble attempt at a solution has been: (1/12)*(1/11)*(1/10), which I know for a fact is wrong. If someone could shed some light, any light, on how to tackle problems like this I would be eternally grateful.

Assuming that a given bulb is not tested more than once (not said in the text, but that seems obvious to me)...

Try to word the problem in a different manner:

Case a) means that your first tested bulb turn out to be a defective one (how many chances for this to happen), same fate for the second one (again, how many chances, given there are only 2 faulty bulbs remaining) as well as for the third (chances?).

Case b) means that your fifth tested bulb turn out to be a defective one, while two out of the four previous ones were also defective. That is:

good, good, bad, bad
or good, bad, good, bad
or good, bad, bad, good
and so on

Try to express all of this in a mathematical fashion and you should see an emerging pattern.

On a), you should have 3/12*2/11*1/10.
(Three chances out of twelve that the first one is defective, then two chances out of 11 for the second to be defective as well, and lastly, one chance out of 11 to get the last defective bulb as the third one.)
Note that this is just according to the binomial formula:
[tex]P=\frac{\binom{3}{3}}{\binom{12}{3}}=\frac{1}{\frac{12!}{3!9!}}=\frac{3!}{12*11*10}=\frac{3}{12}*\frac{2}{11}*\frac{1}{10}[/tex]

Finding the third defective on the fifth attempt means that you have found exactly two defective bulbs in the first four attempts and then on fifth attempt you find the one remaining defective bulb. The last part (finding the one remaining defective bulb on the fifth attempt) should be easy. You could solve the first part by brute force; just enumerate all the possibilities. There is a better way ...

Finding the third defective on the fifth attempt means that you have found exactly two defective bulbs in the first four attempts and then on fifth attempt you find the one remaining defective bulb. The last part (finding the one remaining defective bulb on the fifth attempt) should be easy. You could solve the first part by brute force; just enumerate all the possibilities. There is a better way ...

I see how there must be a better way, but my issue is I don't know that better way, the prof. decided to only give this section about 3 minutes of her time ( -_- ). How do I go about adjusting this? I could go with brute force solve but it wouldn't help me learn this method.

On a), you should have 3/12*2/11*1/10.
(Three chances out of twelve that the first one is defective, then two chances out of 11 for the second to be defective as well, and lastly, one chance out of 11 to get the last defective bulb as the third one.)
Note that this is just according to the binomial formula:
[tex]P=\frac{\binom{3}{3}}{\binom{12}{3}}=\frac{1}{\frac{12!}{3!9!}}=\frac{3!}{12*11*10}=\frac{3}{12}*\frac{2}{11}*\frac{1}{10}[/tex]

So what you're saying DH is that it will lead to a formula that is similar to Arildno's?

In this case, the probability of finding d defective bulbs on or before the tth attempt, [itex]p_{d,t}[/itex], can be defined in terms of [itex]p_{d-1,t-1}[/itex] and [itex]p_{d,t-1}[/itex]. It is not a closed form solution but it is easily computable.

The probability of finding the 3rd defective bulb on the 5th attempt, knowing that you have already found 2 is 1/8, because only 8 bulbs remain, and only 1 of them is defective.

Now...
The probability of finding 2 defective bulbs in the first four attempts is
the probability of the sequence (bad, bad, good, good)
plus the probability of the sequence (bad, good, bad, good)
plus the probability of the sequence (bad, good, good, bad)
etc.

The probability of the sequence B,B,G,G is (3/12)*(2/11)*(9/10)*(8/9)
The probability of the sequence B,G,B,G is (3/12)*(9/11)*(2/10)*(8/9)
etc.

All these elementary probabilities are the same: (3*2*9*8)/(12*11*10*9)
That's because 3*2 represents the decreasing number of faulty bulbs, 9*8 represents the decreasing number of good ones, and 12*11*10*9 represents the decreasing number of total untested bulbs.

Then you only need to figure out the number of ways to sort the sequence B,B,G,G which is 6.

This "6" also comes from the binomial formula, or more accurately from the number of combinations, which in this case is: 4! / (2!(4-2)!)